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robertoCamera

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## Homework Statement

I want to know if there are any conditions we can apply that will give us the following equality:

sum from k=1 to infinity of (-e^(2pi i x))^k = -e^(2pi i x)/ (1+e^(2pi i x))

## Homework Equations

I know in Real Numbers if |r|<1 then we have sum k=1 to infinity of r^k = r/(1-r)

In the Real Numbers things fail when the magnitude is 1. Is the same true in Complex numbers?

## The Attempt at a Solution

We get the equivalence for finite N via induction:

sum k=1 to 1 (-exp(2pi i x))^k=-exp(2pi i x)=(-exp(2pi i x)-exp(2pi i 2x ) ) / ( 1 +exp(2pi i x) )

Then using induction

Assuming

sum k=1 to N (-exp(2pi i x))^k=(-exp(2pi i x)-exp(2pi i (N+1)x ) ) / ( 1 +exp(2pi i x) )

we get

sum k=1 to N+1 (-exp(2pi i x))^k=(-exp(2pi i x)-exp(2pi i (N+1)x ) ) / ( 1 +exp(2pi i x) ) + (-exp(2pi i x))^(N+1)

=(-exp(2pi i x)-exp(2pi i (N+2)x ) ) / ( 1 +exp(2pi i x) )

but since |exp(2pi i x)|=1, we don't get the second term in the numerator going to 0 as N goes to infinity, so what can we do?

Relating to another post:

https://www.physicsforums.com/showthread.php?t=435992

I know that if I could use this equivalence about the geometric sum of complex exponentials then I would be able to say that the derivative of the Fourier series of f(x)=x on [-p/2,p/2) is 1 and thus the Fourier series converges to the function.

I suppose I could just look at this numerically to check, but I would still need some proof that this true in general.Thanks for any help.

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